Does the series diverge or converge

Question

$\sum_{k=1}^\infty\frac{1}{k\ln(k+1)}$

I am unable to determine what method to use to test if this series converges or diverges.

My only clue thus far is that there is a similar problem in our text that uses the integral test to determine that the series diverges.

What else is needed here beyond the integral test to determine that the series diverges?

Thanks

Answer 1

Using Cauchy Condensation test we can consider the convergence of the condensed series $\sum 2^k a_{2^k}$ and we have

$$\frac{2^k}{2^k\ln(2^k+1)}=\frac{1}{\ln(2^k+1)}\sim \frac{1}{k\ln 2}$$

which diverges by limit comparison test with $\sum \frac 1k$.

Answer 2

$$\sum_{k=1}^\infty\frac{1}{k\ln(k+1)} \ge \sum_{k=1}^\infty\frac{1}{(k+1)\ln(k+1)} = \sum_{k=2}^\infty\frac{1}{k\ln(k)} \ge \int_2^\infty \frac 1 {x \ln x} \ \mathrm dx = [\ln \ln x]_2^\infty = \infty$$

Answer 3

$$\sum_{k=1}^\infty\frac{1}{k\ln(k+1)} \ge \sum_{k=1}^\infty\frac{1}{(k+1)\ln(k+1)}$$ The sum on the right diverges by the integral test.

Hence, by the comparison test, the sum on the left also diverges.